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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

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    (Original post by itsConnor_)
    Is my two step mech wrong? If so why? Diff to ms

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    maybe because hydrogen peroxide won't form so they don't accept it i guess?
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    (Original post by itsConnor_)
    Is my two step mech wrong? If so why? Diff to ms

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    Your answer is in the mark scheme.


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    (Original post by itsConnor_)
    Is my two step mech wrong? If so why? Diff to ms

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    The mark scheme will allow that
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    (Original post by suibster)
    Your answer is in the mark scheme.


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    (Original post by vishva97)
    The mark scheme will allow that
    (Original post by lai812matthew)
    maybe because hydrogen peroxide won't form so they don't accept it i guess?
    so sorry guys can't believe i didn't see that. phew

    Liking these 73 = A* grade boundaries (jan 11) lol
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    (Original post by vishva97)
    The mark scheme will allow that
    can N2O also be accepted as one of the species?
    H2+2NO-->N2O+H2O
    H2+N2O-->H2O+N2
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    (Original post by lai812matthew)
    can N2O also be accepted as one of the species?
    H2+2NO-->N2O+H2O
    H2+N2O-->H2O+N2
    yep thats the first thing it says in the ms http://www.thestudentroom.co.uk/atta...2&d=1466247083
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    (Original post by itsConnor_)
    yep thats the first thing it says in the ms http://www.thestudentroom.co.uk/atta...2&d=1466247083
    Hey guys, hope revision is going well.

    Could someone please explaim how to do the following questioms from the June 13 paper?

    The part d) question refers to the table attached.



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    (Original post by zirak46)
    Hey guys, hope revision is going well.

    Could someone please explaim how to do the following questioms from the June 13 paper?

    The part d) question refers to the table attached.



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    You need the rest of the question for the photosynthesis Q (look at the equation and units)
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    Not sure how to go about this.

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    (Original post by zirak46)
    Hey guys, hope revision is going well.

    Could someone please explaim how to do the following questioms from the June 13 paper?

    The part d) question refers to the table attached.



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    Hint for the first Q, DeltaH=E/Mol
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    (Original post by pineneedles)
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    Not sure how to go about this.

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    Okay so it tells you that silver nitrate is used. And all the complexes have chlorine. That means the formula for the precipitate is AgCl. The Mr is 143.3. 2.868 of AgCl is formed. Using moles=mass/Mr you find that it is 0.02 moles which is double 0.01. This means the complex from which it was formed must have 2 chlorine ions. Hope that makes sense.

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    (Original post by ImNervous)
    Okay so it tells you that silver nitrate is used. And all the complexes have chlorine. That means the formula for the precipitate is AgCl. The Mr is 143.3. 2.868 of AgCl is formed. Using moles=mass/Mr you find that it is 0.02 moles which is double 0.01. This means the complex from which it was formed must have 2 chlorine ions. Hope that makes sense.

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    Meaning it would be complex B. The oons with the cobalt will stay with the cobalt.

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    (Original post by ImNervous)
    Meaning it would be complex B. The oons with the cobalt will stay with the cobalt.

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    Yeah, that makes a lot of sense, thank you. That's what was confusing me; I wasn't sure whether the silver nitrate reacted with the chloride ions which are part of the complex or not. Why do they only react with the chloride ions which aren't part of the complex, do you know?

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    (Original post by lai812matthew)
    it specifies dilute which is a bit akward. is 6M dilute? not really imo.
    In the AS textbook it says dilute is anything with a concentration of below 10M.
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    (Original post by pineneedles)
    Yeah, that makes a lot of sense, thank you. That's what was confusing me; I wasn't sure whether the silver nitrate reacted with the chloride ions which are part of the complex or not. Why do they only react with the chloride ions which aren't part of the complex, do you know?

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    No not really
    But I guess it would be because the chloride ions that are part of the complex have formed co - ordinate bonds which are essentially dative covalent bonds and are stronger than ionic bonds. I'm not too sure though. And have you checked the answer on the mrkscheme to make sure its complex B? If not could you please check. Thank you very much
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    Hi, came across a thing about cells that I have never been able to understand.
    In all diagrams, a high-resistance voltmeter is used in the wire between the two electrodes. But electrons need to flow through the wire, from one half cell to another. So why is a high resistance voltmeter used (i.e. why is the flow of electrons impeded)?
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    (Original post by sunsri101)
    Hi, came across a thing about cells that I have never been able to understand.
    In all diagrams, a high-resistance voltmeter is used in the wire between the two electrodes. But electrons need to flow through the wire, from one half cell to another. So why is a high resistance voltmeter used (i.e. why is the flow of electrons impeded)?
    When you are measuring standard electrode potentials you don't want current to flow. As soon as it does the concentrations of solutions start to change.

    V = IR

    So if R is very high, I will be very low.
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    This was actually a redox titration calc from an old thread- just reposting it as I wanted some help with it.

    A solution is made from, 50cm^3 ,iron (ii) chloride, diluted to 500cm^3. the solution was then titrated against both 0.02M potassium manganate, and 0.01677M potassium dichromate (separately). (after adding an excess of dilute sulpuric acid). 25cm^3 solution was used each time, 25.50cm^3 more manganate was needed to be added each time than dichromate. calculate the concentration of iron (ii) chloride in the original solution. before dilution) the answer is 0.51M apparently

    Basically, this is my working so far:
    1) (Cr2O7)2- + 6Fe2+ + 14H+---->6Fe3+ + 2Cr3+ +7H2O
    2) MnO4- +5Fe2+ + 8H+----> 5Fe3+ + Mn2+ +4H2O

    Then, I set: volume of (Cr2O72-)= x, and the volume of (MnO4-)= x+25.5
    for reaction with Cr2O72- moles of Cr2O72-= 0.01677 X (x/1000)= (0.01677x)/1000
    moles of Fe2+ reacted in 25cm^3= 6(0.01677x)/1000= (0.10062x)/1000
    moles of Fe2+ in 500cm^3= 20(0.10062x)/1000= (2.0124x)/1000
    ^^^^This is then the number of moles in 50cm^3 also (before dilution)

    Then for the reaction with MnO4-: moles of MnO4-= 0.02 X (x+25.5)/1000
    moles of Fe2+ reacted in 25cm^3= moles above X5= 0.1(x+25.5)/1000
    moles of Fe2+ in 500cm^3= moles above X 20= 2(x+25.5)/1000
    ^^^^This is also the number of moles in 50cm^3 (before dilution)

    I then equated the two bolded values and came out with like x=4112.9.... Where did I go wrong???
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    Hope this is an easy exam😫
    What were the highest grade boundaries


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    (Original post by TeachChemistry)
    When you are measuring standard electrode potentials you don't want current to flow. As soon as it does the concentrations of solutions start to change.

    V = IR

    So if R is very high, I will be very low.
    But you do want electrons to move between the half cells. Current is the flow of charged particles, so if you want electrons to flow, surely you do want current?
    And the solution concentrations do change over time, that's why batteries run out unless they are rechargeable- right?
 
 
 
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